Probability?

Box A, B and C are identical.





Box A contains 8 identical counters numbered 1,2,3,4,5,6,7 and 8.





Box B contains 6 identical counters numbered 2,3,5,7,8 and 9.





Box C contains 5 identical counters numbered 1,2,3,6 and 8.





A box is chosen at random and a counter is selected at random from the box.





Find the probability that





(a) counter selected has a prime number on it. (ans: 47/90)





Please show your workings clearly...thanks..

Probability?
The probability of selecting a particular box is 1/3 (since there are 3 boxes and you know you are going to select one and only one).





A contains 2, 3, 5 and 7, 4 primes out of 8 numbers. So IF you picked box A, the probability of getting a prime is 1/2. The probability of selecting A and a prime is 1/3*1/2 or 1/6.





Box B has 4 primes out of 6 for 2/3, so B with a prime is 1/3*2/3 or 2/9.





Box C has 2 primes out of 5, so C with a prime is 1/3*2/5 or 2/15.





The probability of picking a prime from any one of the boxes is the sum of the probabilities of picking the prime from a particular box: 1/6+2/9+2/15





1/6 = 135/810


2/9 = 180/810


2/15 = 108/810





The sum of these probabilities is 423/810 = 47/90
Reply:P(selecting a prime|select from Box A)=4/8


P(selecting a prime|select from Box B)=4/6


P(selecting a prime|select from Box C)=2/5





P(selecting a prime)


=P(selecting a prime|select from Box A)*P(select from Box A)


+P(selecting a prime|select from Box B)*P(select from Box B)


+P(selecting a prime|select from Box C)*P(select from Box C)


=4/8*1/3


+4/6*1/3


+2/5*1/3


=47/90
Reply:the possible prime numbers listed from box A are 1, 2, 3, 5, 7 for 5/8 prob of a prime





the possible prime numbers listed from box B are 2,3,5,7 for a 4/6 = 2/3 probability of a prime





the possible prime numbers listed form box C are 1, 2, 3 for a 3/5 probability of a prime





Now, use The Law of Total Probability





For a set of events A1, A2, A3, ... , An where the Ai's are mutually exclusive and exhaustive events and for any other event B





P(B)


= P(B and A1) + P(B and A2) + ... + P(B and An)


= P(B | A1) * P(A1) + P(B | A2) * P(A2) + ... + P(B | An) * P(An)





for this problem





P(prime)


= P(prime | box A) * P(box A) + P(prime | box B) * P(box B) + P(prime | box C) * P(box C)





= 5/8 * 1/3 + 2/3 * 1/3 + 3/5 * 1/3


= 0.6305556